[EM] Proportional preferential voting

[Craig Carey]

I don't appear to have all the equations needed to get a
 complete solution in any 4 candidate problem of my theory.

At 01:55 99/09/17 , you wrote:
>Dear Craig,
>
>could you -please- explain your method using the
>following example with 100 voters and 4 candidates
>running for 3 seats?
>
>30 voters vote A > B > C > D.
>26 voters vote B > D > A > C.
>24 voters vote C > B > A > D.
>20 voters vote D > B > A > C.
>
>Markus Schulze

OK.. here's a rough outline of this method. I am sure it will
 seem defective:

"... solve
 the subproblems, get the rules listed, and when candidate i
 definitely won or lost at some point, cast shadows in every
 direction allowed by the rules throughout the simplex. At some
 points, the number of winners would rise to equal the number
 that had to win the election, and therefore, at each such
 point in the simplex, the losers would all be known. Then the
 shadows for all those losers could be cast.
 The iteration can stabilise without a solution being found."

I could give a sort of answer to the problem from Mr Schulze,
 if the following three 1 winner 4 candidate sub-problems 
 were solved (or partly solved)


VAD
-20 D...
-24 CBA.
-26 BD..
-30 A...
         
VBC
-20 DB..
-24 C...
-26 B...
-30 AB..
         
VBD
-20 D...
-24 CB..
-26 B...
-30 [missing by accident]



Picking a single winner with votes that are negative.

Notation: V is election system
 
Try this notation : aV is true iff A wins V
VA is V with votes after the preference for candidate A discarded.

Then aV = aVA = aVA.-bVAB.-cVAC.-dVAD. That holds for STV & FPP.
 It may hold for Condorcet too? (but not some of the modifications?).

The 3 above because of the solutions I found for VBCD, VACD,...

("-" is Boolean 'not'). 
("CBA." = "C > B > A > D")

------------------------------------
Simplifying concave polytopes would become a problem
 probably. REDLOG seems to be the package to use.

     http://www.fmi.uni-passau.de/~redlog/htmldoc/rl_toc.html

One of the authors of REDLOG wrote:

:Try "rlsimpl((a<b) and (a<c) and (2*a<b+c));",
:but the REDLOG simplifier cannot simplify the formula.
:

[Note: Some inequality term extracting and expression parsing code,
 can add a convex polytope simplification feature.]

:However, with the quantifier elimination of REDLOG you can prove, that
:a<b and a<c implies 2*a<b+c:
:
:	12: rlqe rlall(a<B and a<c impl 2*a<b+c);
:	---- (all a b c) [DFS: depth 3, watching 3]
:	[0e] [DEL:0/1]
:	true
:
:This allows you to drop the atomic formula 2*a<b+c from the above
:conjunction. 
:

G A Craig Carey, 08:46 Fri 17 Sept 1999 NZT







Mr Craig Carey

E-mail: research at ijs.co.nz

Auckland, Nth Island, New Zealand
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